1
$\begingroup$

I am interested in the variational formulation of the 1D Schrodinger equation:

$i u_t- \beta u_{xx} = 0 $ and $u(x,0)=u_0(x)$ which upon integration by parts yields:

$i(u_t,v) + \beta (u_x,v_x) = 0$ and the boundary terms vanish by appropriate testing with test functions $v$. We then have the continuous sesquilinear form $b(u,v) = i(u_t,v) + \beta (u_x,v_x)$. I would like to apply the generalized Lax-Milgram on $b(\cdot,\cdot)$, but I am having trouble showing the boundedness below (coercivity) of $b(\cdot,\cdot)$. Is there a slick way to show $b(u,u) > \alpha \|u\|_{H^1}^2$?

$\endgroup$
15
  • 1
    $\begingroup$ Do you mean space-time $H^1$ in your final inequality? $\endgroup$ Feb 10, 2016 at 15:35
  • $\begingroup$ yes, I meant space-time $\endgroup$ Feb 10, 2016 at 18:12
  • $\begingroup$ or even first with $L^2$ in space-time. I'm having trouble with the complex coefficient $\endgroup$ Feb 10, 2016 at 18:16
  • $\begingroup$ Yes, you are right, I could not control both but I guess then I can control spatial $H^1$ and $L^2$ in time, right? $\endgroup$ Feb 10, 2016 at 18:29
  • $\begingroup$ Let me be a bit less sloppy: does $H^1$ mean $$ \|u\|^2 = \int_t \int_x |u_t|^2 + |u_x|^2 + |u|^2$$ or $$ \|u\|^2 = \int_t \int_x |u_x|^2 + |u|^2$$ or something else altogether? $\endgroup$ Feb 10, 2016 at 18:39

1 Answer 1

3
$\begingroup$

Your method is doomed to fail. For several reasons.

  1. Suppose that $u$ solves the linear Schrodinger equation. Using Fourier methods it is easy to see that $\| u(\cdot,t)\|_{L^2_x}$ is conserved and independent of $t$. This means that $u$ cannot be in $L^2(\mathbb{R}\times \mathbb{R}_+)$. So there is zero chance that Lax-Milgram can give you any indication on how to get a solution.

  2. Forgetting item 1 above. Observe that if Lax-Milgram were to work, you solution $u$ will satisfy $B(u,v) = \langle 0,v\rangle = 0$ for any $v$, since you are solving the homogeneous Schrodinger equation. This implies immediately that $B$ cannot be coercive.

  3. Forgetting items 1 and 2 above, suppose $u$ is a function in $L^2(\mathbb{R}\times\mathbb{R}_+)\cap C^2$, then necessarily $\liminf_{t\to\infty} \|u(\cdot,t)\|_{L^2_x} = 0$. But then unless $u_0 \equiv 0$, $B(u,u)$ must have a non-vanishing imaginary part.

  4. Lax-Milgram is intended to provide a weak solution to the linear partial differential equation $L u = f$ for $u$ belonging to some Hilbert space $H$. In the case $f = 0$ however the existence of a solution is trivial! Namely that $u = 0$ will solve the equation.

    In your problem you are prescribing a boundary value $u_0$. This is not in the usual form of Lax-Milgram. Furthermore, by fixing the boundary value $u_0$ you destroy linearity; in other words, in makes no sense to look at the functional on a Hilbert space $H$, since you cannot add two elements with boundary value $u_0$ and obtain another element with boundary value $u_0$.

$\endgroup$
5
  • $\begingroup$ I agree, but have the following fixes: I could do a weighted $L^2$ norm with a Gaussian weight $e^{-t^2/2}$ so that $\int_{x,t} e^{-t^2/2} |u|^2 = KC_0 < \infty $ and $C_0 = ||u(\cdot,t)||_{L_{x}^2}$ is the conserved total energy. By working with a nonzero right hand side, the issues (2) and (4) regarding homogeneous $u=0$ solution are resolved. $\endgroup$ Feb 10, 2016 at 20:16
  • $\begingroup$ Then for $u$ in your weighted space, $B(u,u)$ cannot possibly be coercive. Let $v\in H^1(\mathbb{R})$ be any function, normalized to have $L^2$ norm 1. Let $\lambda = \| \partial_x v\|_{L^2}$. Now take $u = \exp (i\theta t) v$. Clearly $u$ has finite weighted $L^2_{t,x}$ norm. But note that $i \bar{u}_t u = \theta |v|^2$. Take $\theta < - \lambda$ then you see that $B(u,u)$ cannot be coercive. $\endgroup$ Feb 11, 2016 at 14:58
  • $\begingroup$ Perhaps this is also a good time to mention that for actual solutions to the Schrodinger equation, $\|u_x(\cdot,t)\|_{L^2(\mathbb{R})}$ (the energy) is also conserved in time. So your form $B$ is not continuous there. $\endgroup$ Feb 11, 2016 at 15:10
  • $\begingroup$ One final comment, and no more from me on this topic: many evolution equations in physics can be obtained as a formal critical point of an action functional. The formal Lagrangian for the Schrodinger equation is essentially $B(u,u)$ as you had written. But it is fairly well-known that (1) for most equations of physics the formal Lagrangian does not converge (so really only the Lagrangian density is defined) due to natural conservation laws and (2) that for most equations of physics the solutions are saddle points of the functional (in the sense that the formal second variation has no sign)... $\endgroup$ Feb 11, 2016 at 15:20
  • $\begingroup$ ... both of these make naive space-time variational approaches unsuitable for studying equations of physics. (In contrast, the Hamiltonian approach has much firming functional analytic footing.) The two problems both come up in your attempt to use Lax-Milgram. That for solutions to the linear homogeneous equation $B(u,u) =0$ (formally) and that it is possible to construct functions for which $B(u,u) < 0$ (formally) really is a manifestation of problem (2). $\endgroup$ Feb 11, 2016 at 15:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.